#x=5\lor x=-4#

To see this, we reduce the equation to a quadratic equation:
\[ \begin{array}{rclcl}
{{986\cdot x}\over{7\cdot x-1}}&=&x+140 &\phantom{}&\color{blue}{\text{ given}}\\
986\cdot x &=& (7\cdot x-1)\cdot (x+140) &\phantom{}&\color{blue}{\text{ multiplied by }7\cdot x-1}\\
-7\cdot x^2+7\cdot x+140 &=& 0 &\phantom{}&\color{blue}{\text{ all terms to the left hand side }}\\
x^2-x-20 &=& 0 &\phantom{}&\color{blue}{\text{ divided by }-7}\\
\end {array} \]
The solutions of this quadratic equation are: #x=5\lor x=-4#.
Because neither of the two is equal to #x=\frac{1}{7}# (in which case the denominator in the original equation would be equal to zero), they are also solutions to the original equation.